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ANSWERS KEY

CHEMINSTRY

1d

2b

3c

4b

5b

6b

7c

8b

9a

10c

11d

12d

13 c 14 a 15 b 16 c 17 d 18 d 19 c 20 c 21 b 22 d 23 a 24 c

25 b 26 c 27 b 28 c 29 c 30 c

PHYSICS

1 A 2 a 3 b 4 a 5 c 6 d 7 d 8 d 9 c 10 d 11 b 12 b

13 a 14 b 15 d 16 a 17 a 18 b 19 b 20 c 21 b 22 a 23 d 24 c

25a

26c

27b

28d

29d

30A

MATHEMATICS

1 a 2 b 3 b 4 d 5 a 6 a 7 b 8 b 9 a 10 d 11 b 12 d

13 b 14 b 15 c 16 c 17 d 18 b 19 c 20 a 21 a 22 d 23 a 24 b

25 a 26 c 27 b 28 c 29 a 30 c

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CHEMINSTY

Sol 1 .

Hot alkaline KMnO 4 oxidizes I

– bytyne to CH 3CH2COOH and CO2.

H3C

– CH2

–

C ≡ CH KMnO

Heat H3C

– CH2

– COOH + CO2

Sol 2 .

Benzoyl chloride is reduced to benzaldehyde

(Rosenmund’s reduc

Sol 3 .

Since rate increases 2 times on increasing the initial concentration of reactant 4 times, rate dependsupon ½ power of the concentration, i.e., Rate = k [A] 1/2 , these order of reaction is 1/2.

Sol 4 .

Q = I x t = 2 x 20 x 60 (s) = 2400 C

2H2O → O2 + 4 H + + 4 e4 x 96500 C are required to O 2 =

22.4

4 96500 x 2400 = 0.1392 L

Sol 5 .Cyclobutadiene is electrons in a cyclic conjugated system. Benzene, o-xylene andpieric acid are aromaaromaticity.

Sol 6 .

Cu2+ , Fe2+ and Fe 3+ complexes are colored because of presence of unpaired d – electron;Cd2+ complex is colored because all electrons in d-subshell (d 10 case) are paired.

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Sol 7 .

The enthalpy change depends upon the final and initial state of the reaction; the intermediates donot effect overall enthalpy change of a reaction

Sol 8 .

For an ideal gas, compressibility factor, Z = = 1

Sol 9 .

Calcium can be obtained by electrolysis of molten CaCI 2

Sol 10 .

Alkynes react with sodium metal In liquid ammonia to form trans alkenes.

Sol 11

CaC2 → C2H2 → C2H2 → polyethene 64 26 28 n x 28

64 kg 26 kg 28 kg 28 kg

Thus as per balanced equation. Amount of polyethene obtained from 64 kg of clcium carbide is 28kg.

Sol 12

Nesseler’s reagent K2 [HgI4] is used for detection of ammonium ion. With ammonia it gives areddish brown ppt or colouration.Sol 13

Cu2+ + 2I -

→ Cu + I2

Sol 14

Molecular orbital energy level diagram for O 2 shows two unpaired electrons in antibonding pi-molecular orbitals so it is paramagnetic.

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Sol 15

For pH = 1, [H + ] = 10 -1; for pH = 2, [H + ] = 10 -2

Total [H + ] = 0.1 + 0.1 = 0.11

pH = - log (0.11) = 0.9586

Sol 16

Dehalogenation of 1,3 – dibromopropane with Zn givscyclopropane.Sol 17

Hydrolysis of CH 3CN gives acetic acid and ammonia.

CH3CN CH3COOH + NH 3

Sol 18

Total number of stereo-isomers of CH 3CHBrCHBrCOOH is 4 and for CH 3CHBrCH is 3.

`

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Sol 19

Lithium is less electronegative as compared to hydrogen so hydrogen bonded to lithium in LiAlH 4 istransferred as hydride ion.

20/IUPAC name of the given compound is:

Sol 21

I2 + I -→ 3−, I- gives electrons to I 2, it acts as a Lewis base.Sol 22

On heating in a charcoal cavity Zn 2+ salt formsZnO which on combination with CoO (fromCo(NO3) 2) gives a green mass CoO. ZnO.

Sol 23

R- CH2CH2OH can be coverted into R – CH2CH2COOH with the help of PBr 3, KCN, H+

R

– CH2-CH2-OH

3 R-CH2-CH2-Br R-CH2-CH2CN

+

R-CH2-CH2-COOH

Sol 24

Al (OH) 3 dissolves in excess of NaOH solution to form aluminates.

Al(OH) 3 + 3NaOH → Na3AlO3 + 3H 2 OSol 25

All six bonds are not similar in diborane.

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Sol 26

In SiO2, silicon is sp 3 – hybridized.

Sol 27

PF3 can undergo rapid hydrolysis

2PF 3+ 3h 2

O → 2HF + H3PO4 + HPF 4

Sol 28

CO is a neutral ligand therefore, oxidation state of Ni in [Ni(CO) 4] is zero.

Sol 29

Iodo form is formed.

Sol 30

The role of acetic anhydride which is used as a solvent in CrO 3 oxidation of toluene to benzaldehydeis to protect further oxidation of benzaldehyde.

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PHYSICS

Sol 1

Here [X] =2

−2 2

−1 2

−1 3 −2 2 5 =

[ 3 6 −4[ 3 6 −4 ] = [ 0 0 0 ]∴xis a dimension less quantity⇒xmust be an angle is a dimensionless quantitySol 2

Magnitudes and directions of velocity and accelerationare individual and to be seen in this contextalso

Sol 3

D’ Alembert and IneSol 4

For glass balls e = 0.94

For ivary balls e = 0.81

For cork balls e = 0.65

For lead balls e = 0.20

Coefficient of restitution is maximum for glass balls.

Sol 5

A ball at rest on a horizontal surface has neutral equilibrium

Sol 6

In an artificial satellite g becomes meaningless due to free fall of spaceship which causes a feeling ofweightlessness.

Sol 7

As viscosity has high values at low temperatures, the cold fluid flows sluggishly

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Sol 8

Neutral temperature is independent of the temperature is independent ofhot junction, temperatureof cold junction and temperature of inversion

Sol 9

All the molecules in a mixture of ideal gases have the same mean translational kinetic energy

Sol 10

Using T = 2π As the acceleration due to gravity becomes (5 g + g ) = 6g, as mentioned in the question, we getnew time period as

T’ = 2π 6 = 6

Sol 11

The time periods for both bodies will be same as T 1 = T 2 = 2π where R is the radius of the earthdecreases.Sol 12

As the dielectric is introduced, charge on plate decreases.

Sol 13

The formula The formula for drawing power at a voltage V is P =0

0

Sol 14

The ratio of radii

=

Sol 15.

The susceptibility of ferromagnetic substances strongly depend on temperature and strength ofmagnetic field

Sol 16

Statement 1 and 2 are self explanatory

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Sol 17

The acceleration of magnet M will be larger than g when it is below the ring R and moving awayfrom it.

Sol 18

Magnetic and electric vectors in an electromagnetic oscillations are in same phase where as there isdifference of π/2 in t Sol 19

Using d =sin ′ we get

D =400 100−9

sin 30 0 =

400 10−91/2

= 800 x 10 -7 m

Again using d sin θ 3

λ2d

Sin θ =3 4 10−72 8 10−7 = 34 ⇒θ = sin -1 34 Sol 20

Without mirror:

Using Lens formula1 – 1 = 1

Given u = - 12 cm, = 10 cm⇒1 + 112 = 110 ⇒

1 =

1

10

–

1

12 =

1

60

⇒ v = 60 cm

Which shows that in the absence of mirror, the image is formed at 60 cm away from the lens.

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Sol 21

With Mirror :

Using above figure

PI1 = radius of curvature of the mirro

i.e. R = 60 – 10 = 50 cm⇒ = 2 =

50

2 = 25 cm

Sol 22

Kinetic energy of electron is greater than kinetic energy of proton due to larger velocities.

Sol 23

The infrared radiation will be emitted if the transition between the staten = 5 to n = 4 is allowedsince the infrared radiation has less energy than that of ultraviolet radiation.

Sol 24

After 2 days 1000 atoms become 100.

After another two days 100 atoms will remain100

10 i.e.e 10 atoms

∴

After 4 days 10 atoms will remain.

Sol 25

Since pentavalent impurity atom contributes one electron to the intrinsic semiconductor, 10 24 impurity atoms per unit volume will contribute 10 24 electrons to the intrinsic semiconductors.Therefore, the free electron density will increase by 10 24 m -3

Sol 26

Satellite communication and line of sight (LOS) communications (e.g. MW radio waves) are carriedthrough space waves

Given d 1 =

2 1 and d 2 =

2 2 For maximum range

Dmax = d 1 + d 2 = d D max = 2 1 + 2 2 i.e. d 1 = d 2 or 2 1 = 2 2 i.e. h 1 = h 2 If h1 + h 2 = h

Then h 1 = 2

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Sol 27.

Givne2

∝ R-5/2

⇒

v

∝

R3/4

As T =2

⇒ T2∝2

or T 2∝−342

or T 2∝ R7/2 Sol 28

Mass of hanging portion is3

and centre of mass C is at a distance h = L/6 below the table top.

∴the required workdone, W = mgh=

3

6=

18

Sol 29

The centre of mass will always remain at rest as net force on centre of mass zero.

Sol 30

In uniform circular motion, centripetal force acting on the particle. The torque due to this forceabout the centre is zero. Hence angular momentum about centre remains conserved.

MATHEMATICS

Sol 1 Given Re+

+ = 1

⇒

( + )+ +

++

2 = 1

⇒ (az + b) (c + d) + (a + b) (cz + d) = 2 (cz + c) (c + d)

⇒bcz + bd = 2 (c 2 z + cdz + cd + d 2)

⇒2ac +ad (z + ) + bc (z + )+ 2bd = 2 [c 2 z + cd (z + ) + d 2]⇒ (2ac – 2c 2) (x 2 + y 2) + (ad + bc – 2cd) (2x) + 2 (bd – d2) = 0It is a circle.

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Sol 2

Given that

+

− (1

−) = 1 (i)

− (1 −) = 1 – Squaring both sides, we getx– (1 −) = 1− 2 x– (1−) =1 −2 + xx– (1 −) =1 −2 Squaring again on both sides, we get1– x = 1 + 4x – 4 ⇒4 = 5xSquaring on both sides again, we get

16 x = 25 x 2

⇒ 25 x 2 = 16x = 0

⇒x (25x – 16) = 0⇒ x = 0, x = 1625 x = 0 does not satisfy (i)

Therefore x =16

25 So (i) has only one real root.

Sol 3

Given sequences are 2, 4, 6, 8, 10, 12, 14, 16 and 1, 4, 7, 10, 13, 16

Common terms are 4, 10, 16 . . . . . . . . . . . . . . . .

n th terms of the common sequence

= 4 + (n - 1) (6) = 6n – 2100 th term of the first term sequence

= 2 + (100 - 1) 2 = 200

100 th term of the second sequence

= 1 + (100 - 1) 3 = 300

⇒t n ≤ 200 ⇒ 6n– 2≤ 200⇒ 6n≤ 198⇒n≤ 33 ⇒ b = 23

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Sol 4

For positive integral solutions

36 = 36 x 1x 1 = 18 x 2 x 1 = 12 x 3x 1 = 9 x 4 x 1 = 9 x 2 = 6x 6 x 1 = 4 x 3 x 3

Number of positive integral solutions

=3!

2! +

3!

1! +

3!

1! +

3!

1! +

3!

2! +

3!

2! +

3!

2!

= 3 + 6 + 6 + 6 + 3 + 3 + 3

= 30

Sol 5

720

– 5 20 = (7

– 5 )(7 19 + 7 18 . 51 + 7 17 . 52 + . . . . . . . . . . . . . . . . . . . + 519 )

∴ 720

– 520 is divisible by 2

Sol 6

∆1 = 1 1 12 2 2 = (a - b) (c – c ) (c -a)∆2 = 1 1 12 2 2 ∆3 = 1 1 1 = 2

2 2 2 = 2 1 1 1

2 2 2 = - 2

1 1 1

2 2 2

∴ 2∆1 + ∆2 = 0Sol 7

If A is orthogonal matrix then AA T = I = A T A

Sol 8

Let x = log 20 3 =log 3

log 20 =

1log 27

3log 20>

1

3

∵log 27

log 20> 1

∴1

3< x <

1

2

∴x∈1

3,

1

2

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Sol 9

Let ABCDEF be a regular hexagon.

Total number of triangles = 6 3

Total form the equilateral the

Vertices should be either A, C, E or B, D, F

∴ Number of equilateral triangles = 2The required probability =

2

20 =

1

10

Sol 10

Let E donates the event that sum of eight occurs on the die and A the event that the man reports

that it is eight. The events that sum of eight occurs are {(2, 6), (3,5), (4,4), (5,3), (6,2)}

P(E) =5

36 P (E’) =3136 P (A/E) = 75

100 =

3

4& P (A/E’) =14 By Baye’s TP (E/A) = ()( / )()( / ) + (′)( / ′)=

536

34

536

34 +

3136

14

P (E/ A) =15

14415

144 +31

144

P (E/ A) = 1546

Sol 11

Given that

(x) = 1 + cos 2x + cos 4 + . . . . . . . . . . . (x) = 11−cos 2 = 1sin 2 ′

(x) = -sin

3 > 0

∵

x

∈−2

, 0

⇒ (x) is increasing function

∴ Range = −2 , (0) ∴( →−2 (), →0 ( )) = (1, ∞)

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Sol 12

Let 1 (x) is defined when – 1 ≤1−| |3 -

3 ≤ 1-

|x| ≤ 3

-

4 ≤ 1-

|x| ≤ 2

-2 ≤ |x| ≤ 2 ⇒ - 4≤ x≤ 4 f2= [x] is defined for every real number

Df2 = R

We know D f = D f1

∩ Df2

∴Df = [- 4, 4]

Sol 13

Given that

(x) = 12 ( −1)+ 3 ( −); (x) > 0 ∀ x ϵ R

Let Lt

x→∞ () = c

∴ Lt x→∞

(x - 1) = c

andLt x → ∞ (x - 1) = c∴ c =12 + 3 2c 2 = c 2 + 3

⇒ c2 = 3⇒ c = ±

3

∴ c = 3∴ Ltx → ∞ ( ) = 3

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Sol 14

Given that

(){

−1

()−2

1

≠2

=2

→2− ()= →2 cos−1 ( )−2Put x =

2 - h

Then x →2 − →0→2−

()= →0 cos−

1

2

−

−

= →0 cos −1 (cos )− = →0 - = - 1

→2 + ()= →0 cos−1 2 + = →

0

cos

−1 (cos )

= →0 = 1∴ →2− ()≠ →2 + ( )∴ →2 () Sol 15

The given function is f (x) = |x - 1| + cos |x|

()= − ( −1)+ , < 0− ( −1)+ , 0 ≤< 1( −1)+ , 1 < 1 ()= − 1−, < 0− 1−, 0 ≤< 11−, 1 < It is clear that f (x) is not differentiable at x = 1

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Sol 16

Given that ++ + …..∞

∴ y = ex + y

log e y = (x +y)

Differentiating both sides w.r.t.x, we get

1= 1 +

1−1 = 1 =

1− Sol 17

Given that

− – + = c Differentially both sides w.r.t.x, we get12( −)−12 1−−12(+ )−12 1 + = 0

1

2 − – 12 + = 12 − + 12 + + − −2 + − = + + −2 + − =

+

–

−

+ + − = − + + − ( , ) = − 2 Sol 18

Given T = 2π Taking log on both sides

logT log2π +1

2 (log l

– log g)

Taking differential on both sides

=1

2 − x 100 =

1

2 100 − 100

=12 (2– 1.5) = 0.52 = 0.25

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Sol 19

(x) = ∫ e2x (x - 4) (x - 5) dx ′

(x) = e 2x (x - 4) (x - 5) < 0

If (x - 4) (x - 5) < 0

If 4 < x < 5

Sol 20

Given that (x) = e xCosx ′ (x) = e xcosx– exsinx

′ (x) =

2

1

2

−1

2

=

2 sin

4

−cos

4

= 2 sin 4− f" (x) = 2 4− − 2 4− = 2e x

1 2 4− −1 2 cos 4− = 2e x

4

− cos π4

− cos

4

− sin

4

= 2 e x sin4− −4 = 2e x sin ( - x) = -2e xSinxFor maximum slope

⇒ sinx = 0 ⇒ x = 0, f" (x) = 2 e x sin

2− + 2e x Cos 2− f” (x) =

- 2ex

Sinx

– 2 e

x

cosx

= - 2 2 1 2 + 1 2 = - 2 2 sin + 4 f" (0) = - 2 2 sin 4 = - 2 < 0 Maximum slope at x = 0

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Sol 21

Let I = ∫(1+ )1/2 − (1+ )1/3 Put 1 + x = t 6

⇒ dx = 6t5dt

I = ∫6 5

( 3− 2 ) = 6 ∫3

−1dt= 6 ∫3− 1 + 1( −1) dt= 6 ∫ 2 + + 1 + 1 −1 = 6

3

3+

2

2+ + 1 | −1| + c

I = 2 (1 + x ) 1/2 + 3 (1 + x ) 1/3 + 6 (1 + x) 1/6

+ 6ln (1 + )16 −1 + cSol 22I = ∫10 2 cot −1 1−1+ Put x = cos θ ⇒ dx =-sin θ do I =

∫0

/22cot

−1

1

−cos

1+cos

(−sin

)

= ∫cos(2 cot −1 tan /2)20 sin d = ∫ cos 2 2−2/20 sin θ d θ = ∫ cos( −)/20 sin θ d θ =

∫−/2

0

sin θ d θ

=cos 2

θ2 0/2

=1

2cos 2

2 –cos 2 0

I =1

2(0−1)= −12

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Sol 23

Given = ( −)( −)

⇒ x2lnxdy

– xydy = xy dx

– y2 lny dx

Dividing both sides by x 2 y2, we get

2 dy– 1 dy = 1 dx– 2 dx −12 + 1 + −12 + 1 = 0

D + = 0

Integrating both sides, we get

+ =c

⇒ +

=

Sol 24

Given bisector is x + 2y – 5 = 0Since the bisectors are perpendicular to each other

∴ other bisector is 2 y = 0

Sol 25

Given equation of circle is

X2 + y 2 – 2x– 4y– 4 = 0 having centre (1,2) and radius = 3If x = λ is parallel to2− 1 = ± 3⇒ 2 – λ = ± 3 ⇒ λ = 2 ± 3

⇒ λ = 5,- 1Pair of lines (x - 5) (x - 1) = 0⇒ x2 – 4x– 5 = 0

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Sol 26

Let P (2t 2, 4t) be any point in the parasols y 2 = 8x

Normal at point p on y 2 = 8x

y + tx = 4t + 2t 3

It must passes through centre of circle (0, - 6) for minimum distance between the thro curves.

- 6 + 0 = 4t + 2t 3

t 3 + 2t + 3 = 0

t = - 1

∴ Point is P (2, 4)

Sol 27

Equation of tangent of y 2 = 4ax in terms of slope is y = mx + ′ where m is slope of tangent.This is also tangent of

2

2 +2

2 = 1

∴2 = a 2 m2+b 2

⇒ a2 12− 2 = 2 ⇒

1

−4

2 =2

2

⇒1+ 2 (1−2 )2 = 22 ⇒

1−22 = 22 (1+ 2 )> 0⇒

1−22 > 0

⇒

2−12 < 0⇒ 0 < m

2< 1

⇒ m ∈ (-1, 0) ∪ (0, 1) ⇒ m ∈ (0, 1) for po

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Sol 28

Equation of plane through (1, 2, 3) is

a (x - 1) + b (y - 2) + c (z - 3) = 0 (i)

Which is perpendicular to

x + 2y + 4z = 1andx – 3y– 5z = 2∴ a + 2b + 4c = 1 (ii)A– 3b– 5c = 2 (iii)Eliminating a, b, c from (i), (ii) and (iii) we get

−1

−2

−3

1 2 4

1

−3

−5

= 0

i.e. 2 (x - 1) + 9 (y - 2) – 5 (z - 3) = 02x + 9y – 5z– 5 = 0Sol 29

2 + 2 + 2 + …+ 2(1 −)(n number of 2 + 2 + 2 + …+ 2 + 2 s i n 2 ((n -1)number of

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

= 2 + 2 s i n 2 −1 =

2 1 + 2 s i n 2 2

− 1

= 2 sin2

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Sol 30

Given cotx + cosecx = 2 sin x

Multiply both sides bcosx + 1 = 2 sin 2 x

cosx + 1 = 2 (1 – cos 2 x)cosx + 1 = 2 (1 - cosx) (1 + cosx)(1 + cosx) – 2 (1 - cosx) (1 + cosx) = 0(1 + cosx) [1 – 2 (1 - cosx)] = 0(1 + cosx) (- 1 + 2cosx) = 0

cosx = -1, cosx =1

2

But cosx ≠-1 as sinx ≠ 0 ∴ cosx =12 ⇒ x =3 , 53 Number of solution in the interval [0,2 π] are 2